- •24.3 HYDRAULICS
- •24.4 OTHER SYSTEMS
- •24.5 SUMMARY
- •24.6 PRACTICE PROBLEMS
- •24.7 PRACTICE PROBLEM SOLUTIONS
- •24.8 ASSIGNMENT PROBLEMS
- •25. CONTINUOUS CONTROL
- •25.1 INTRODUCTION
- •25.2 CONTROL OF LOGICAL ACTUATOR SYSTEMS
- •25.3 CONTROL OF CONTINUOUS ACTUATOR SYSTEMS
- •25.3.1 Block Diagrams
- •25.3.2 Feedback Control Systems
- •25.3.3 Proportional Controllers
- •25.3.4 PID Control Systems
- •25.4 DESIGN CASES
- •25.4.1 Oven Temperature Control
- •25.4.2 Water Tank Level Control
- •25.5 SUMMARY
- •25.6 PRACTICE PROBLEMS
- •25.7 PRACTICE PROBLEM SOLUTIONS
- •25.8 ASSIGNMENT PROBLEMS
- •26. FUZZY LOGIC
- •26.1 INTRODUCTION
- •26.2 COMMERCIAL CONTROLLERS
- •26.3 REFERENCES
- •26.4 SUMMARY
- •26.5 PRACTICE PROBLEMS
- •26.6 PRACTICE PROBLEM SOLUTIONS
- •26.7 ASSIGNMENT PROBLEMS
- •27. SERIAL COMMUNICATION
- •27.1 INTRODUCTION
- •27.2 SERIAL COMMUNICATIONS
- •27.2.1.1 - ASCII Functions
- •27.3 PARALLEL COMMUNICATIONS
- •27.4 DESIGN CASES
- •27.4.1 PLC Interface To a Robot
- •27.5 SUMMARY
- •27.6 PRACTICE PROBLEMS
- •27.7 PRACTICE PROBLEM SOLUTIONS
- •27.8 ASSIGNMENT PROBLEMS
- •28. NETWORKING
- •28.1 INTRODUCTION
- •28.1.1 Topology
- •28.1.2 OSI Network Model
- •28.1.3 Networking Hardware
- •28.1.4 Control Network Issues
- •28.2 NETWORK STANDARDS
- •28.2.1 Devicenet
- •28.2.2 CANbus
- •28.2.3 Controlnet
- •28.2.4 Ethernet
- •28.2.5 Profibus
- •28.2.6 Sercos
- •28.3 PROPRIETARY NETWORKS
- •28.3.1 Data Highway
- •28.4 NETWORK COMPARISONS
- •28.5 DESIGN CASES
- •28.5.1 Devicenet
- •28.6 SUMMARY
- •28.7 PRACTICE PROBLEMS
- •28.8 PRACTICE PROBLEM SOLUTIONS
- •28.9 ASSIGNMENT PROBLEMS
- •29. INTERNET
- •29.1 INTRODUCTION
- •29.1.1 Computer Addresses
- •29.1.2 Phone Lines
- •29.1.3 Mail Transfer Protocols
- •29.1.4 FTP - File Transfer Protocol
- •29.1.5 HTTP - Hypertext Transfer Protocol
- •29.1.6 Novell
- •29.1.7 Security
- •29.1.7.1 - Firewall
- •29.1.7.2 - IP Masquerading
- •29.1.8 HTML - Hyper Text Markup Language
- •29.1.9 URLs
- •29.1.10 Encryption
- •29.1.11 Compression
- •29.1.12 Clients and Servers
- •29.1.13 Java
- •29.1.14 Javascript
- •29.1.16 ActiveX
- •29.1.17 Graphics
- •29.2 DESIGN CASES
- •29.2.1 Remote Monitoring System
- •29.3 SUMMARY
- •29.4 PRACTICE PROBLEMS
- •29.5 PRACTICE PROBLEM SOLUTIONS
- •29.6 ASSIGNMENT PROBLEMS
- •30. HUMAN MACHINE INTERFACES (HMI)
- •30.1 INTRODUCTION
- •30.2 HMI/MMI DESIGN
- •30.3 DESIGN CASES
- •30.4 SUMMARY
- •30.5 PRACTICE PROBLEMS
- •30.6 PRACTICE PROBLEM SOLUTIONS
- •30.7 ASSIGNMENT PROBLEMS
- •31. ELECTRICAL DESIGN AND CONSTRUCTION
- •31.1 INTRODUCTION
- •31.2 ELECTRICAL WIRING DIAGRAMS
- •31.2.1 Selecting Voltages
- •31.2.2 Grounding
- •31.2.3 Wiring
- •31.2.4 Suppressors
- •31.2.5 PLC Enclosures
- •31.2.6 Wire and Cable Grouping
- •31.3 FAIL-SAFE DESIGN
- •31.4 SAFETY RULES SUMMARY
- •31.5 REFERENCES
- •31.6 SUMMARY
- •31.7 PRACTICE PROBLEMS
- •31.8 PRACTICE PROBLEM SOLUTIONS
- •31.9 ASSIGNMENT PROBLEMS
- •32. SOFTWARE ENGINEERING
- •32.1 INTRODUCTION
- •32.1.1 Fail Safe Design
- •32.2 DEBUGGING
- •32.2.1 Troubleshooting
- •32.2.2 Forcing
- •32.3 PROCESS MODELLING
- •32.4 PROGRAMMING FOR LARGE SYSTEMS
- •32.4.1 Developing a Program Structure
- •32.4.2 Program Verification and Simulation
- •32.5 DOCUMENTATION
- •32.6 COMMISIONING
- •32.7 REFERENCES
- •32.8 SUMMARY
- •32.9 PRACTICE PROBLEMS
- •32.10 PRACTICE PROBLEM SOLUTIONS
- •32.11 ASSIGNMENT PROBLEMS
- •33. SELECTING A PLC
- •33.1 INTRODUCTION
- •33.2 SPECIAL I/O MODULES
- •33.3 SUMMARY
- •33.4 PRACTICE PROBLEMS
- •33.5 PRACTICE PROBLEM SOLUTIONS
- •33.6 ASSIGNMENT PROBLEMS
- •34. FUNCTION REFERENCE
- •34.1 FUNCTION DESCRIPTIONS
- •34.1.1 General Functions
- •34.1.2 Program Control
- •34.1.3 Timers and Counters
- •34.1.4 Compare
- •34.1.5 Calculation and Conversion
- •34.1.6 Logical
- •34.1.7 Move
- •34.1.8 File
- •34.1.10 Program Control
- •34.1.11 Advanced Input/Output
- •34.1.12 String
- •34.2 DATA TYPES
continuous actuators - 24.23
ef = rfif + lfifD |
|
|
|
|
|
||||||||
T = KTif |
|
|
|
|
|
|
|
|
|
||||
T |
|
|
2 |
|
|
|
|
|
|
|
|
|
|
--θ |
= |
JD |
|
|
+ BD |
|
|
|
|
|
|||
θ |
= |
|
|
|
1 |
|
|
|
|
|
|
|
|
-- |
----------------------- |
|
|
|
|
|
|||||||
T |
|
JD |
2 |
+ BD |
|
|
|
|
|
||||
|
|
|
|
|
|
|
|
||||||
θ |
= |
θ T |
|
|
= |
|
|
|
KT |
|
|
||
-- |
-- -- |
|
|
----------------------- |
|
|
|||||||
if |
|
T if |
|
|
|
JD2 + BD |
|
|
|||||
θ |
= |
θ if |
|
= |
|
|
|
|
KT |
|
1 |
||
--- |
-- --- |
----------------------- |
----------------- |
||||||||||
ef |
|
if ef |
|
|
JD |
2 |
|
|
rf + lfD |
||||
|
|
T if |
|
|
|
+ BD |
|
||||||
T |
= |
= |
K |
|
|
|
1 |
|
|
||||
--- |
-- --- |
T |
----------------- |
||||||||||
ef |
|
if ef |
|
|
|
|
|
|
|
||||
|
|
|
|
|
|
rf + lfD |
|
Figure 24.20 Equations for a controlled field motor
24.3 HYDRAULICS
Hydraulic systems are used in applications requiring a large amount of force and slow speeds. When used for continuous actuation they are mainly used with position feedback. An example system is shown in Figure 24.21. The controller examines the position of the hydraulic system, and drivers a servo valve. This controls the flow of fluid to the actuator. The remainder of the provides the hydraulic power to drive the system.
continuous actuators - 24.24
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
controller |
|
|
|
|
valve |
|
|
hydraulic |
|||
|
|
|
|
|
|
|
power |
|||||
|
|
|
|
|
|
|
|
|
||||
|
|
|
|
|
|
|
|
|||||
|
|
|
|
|
|
|
|
|
|
|
supply |
|
|
|
|
|
position |
|
|
|
|
|
|||
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
position |
|
|
|
|
||||||
|
|
|
|
|
hydraulic |
|
|
|
||||
|
|
sensor |
|
|
|
|
|
|
||||
|
|
|
|
|
|
|
|
|||||
|
|
|
|
|
actuator |
|
|
|
||||
|
|
|
|
|
|
|
|
|
|
|||
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
sump
Figure 24.21 Hydraulic Servo System
The valve used in a hydraulic system is typically a solenoid controlled valve that is simply opened or closed. Newer, more expensive, valve designs use a scheme like pulse with modulation (PWM) which open/close the valve quickly to adjust the flow rate.
24.4 OTHER SYSTEMS
The continuous actuators discussed earlier in the chapter are the more common types. For the purposes of completeness additional actuators are listed and described briefly below.
Heaters - to control a heater with a continuous temperature a PWM scheme can be used to limit a DC voltage, or an SCR can be used to supply part of an AC waveform.
Pneumatics - air controlled systems can be used for positioning with suitable feedback. Velocities can also be controlled using fast acting valves.
Linear Motors - a linear motor works on the same principles as a normal rotary motor. The primary difference is that they have a limited travel and their cost is typically much higher than other linear actuators.
Ball Screws - rotation is converted to linear motion using balls screws. These are low friction screws that drive nuts filled with ball bearings. These are normally used with slides to bear mechanical loads.
continuous actuators - 24.25
24.5SUMMARY
•AC motors work at higher speeds
•DC motors work over a range of speeds
•Motion control introduces velocity and acceleration limits to servo control
•Hydraulics make positioning easy
24.6PRACTICE PROBLEMS
1.A stepping motor is to be used to drive each of the three linear axes of a cartesian coordinate robot. The motor output shaft will be connected to a screw thread with a screw pitch of 0.125”. It is desired that the control resolution of each of the axes be 0.025”
a)to achieve this control resolution how many step angles are required on the stepper motor?
b)What is the corresponding step angle?
c)Determine the pulse rate that will be required to drive a given joint at a velocity of 3.0”/sec.
2.For the stepper motor in the previous question, a pulse train is to be generated by the robot controller.
a)How many pulses are required to rotate the motor through three complete revolutions?
b)If it is desired to rotate the motor at a speed of 25 rev/min, what pulse rate must be generated by the robot controller?
3.Explain the differences between stepper motors, variable frequency induction motors and DC motors using tables.
continuous actuators - 24.26
24.7 PRACTICE PROBLEM SOLUTIONS
1.
|
|
in |
|
in |
|
|
|
|
|
|
a) |
P = 0.125 rot------- |
R = 0.025step---------- |
|
|
|
|
|
|
||
|
|
|
in |
|
|
|
|
|
|
|
θ |
R |
0.025step---------- |
rot |
Thus |
|
1 |
= |
5 |
step |
|
= -- |
= -------------------------- = |
0.2---------- |
------------------ |
rot |
----------rot |
|||||
|
P |
in |
step |
|
0.2 |
|
|
|||
|
|
0.125 -------rot |
|
|
|
step |
|
|
|
|
b) θ |
|
rot |
deg |
|
|
|
|
|
|
|
|
---------- |
---------- |
|
|
|
|
|
|
|
|
= 0.2step = 72step |
|
|
|
|
|
|
|
|||
c) |
|
3in---- |
= 120steps |
|
|
|
|
|
|
|
PPS = |
------------------------s |
|
|
|
|
|
|
|||
|
|
0.025----------in |
|
s |
|
|
|
|
|
|
|
|
step |
|
|
|
|
|
|
|
2.
a) |
|
|
|
|
|
step |
|
|
|
|
|
|
|
|
|
||
|
pulses |
|
|
|
|
---------- |
= 15steps |
|
|
|
|
|
|
||||
|
= ( 3rot) 5 rot |
|
|
|
|
|
|
||||||||||
b) |
pulses |
= |
|
25 |
rot |
|
5 |
step |
|
= 125 |
steps |
= 125 |
|
1min |
steps |
= 2.08 |
step |
----------------s |
--------- |
----------rot |
------------min |
------------ |
------------ |
----------s |
|||||||||||
|
|
|
|
min |
|
|
|
|
|
60s |
min |
|
3.
|
speed |
torque |
|
|
|
|
|
stepper motor |
very low speeds |
low torque |
|
vfd |
limited speed range |
good at rated speed |
|
dc motor |
wide range |
decreases at higher speeds |
24.8 ASSIGNMENT PROBLEMS
1.A stepper motor is to be used to actuate one joint of a robot arm in a light duty pick and place application. The step angle of the motor is 10 degrees. For each pulse received from the pulse train source the motor rotates through a distance of one step angle.
a)What is the resolution of the stepper motor?
b)Relate this value to the definitions of control resolution, spatial resolution, and accuracy, as discussed in class.
c)For the stepper motor, a pulse train is to be generated by a motion controller.
continuous actuators - 24.27
How many pulses are required to rotate the motor through three complete revolutions? If it is desired to rotate the motor at a speed of 25 rev/min, what pulse rate must be generated by the robot controller?
2.Describe the voltage ripple that would occur when using a permanent magnet DC motor as a tachometer. Hint: consider the use of the commutator to switch the polarity of the coil.
3.Compare the advantages/disadvantages of DC permanent magnet motors and AC induction motors.